Intermediate Math Problem of the Week 9/14/2017

[PF PotW Robot]

Here is this week's intermediate math problem of the week. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Occasionally there will be prizes for extraordinary or clever methods.

Let $f : [a,\infty)\to \Bbb R$ be a continuous function that satisfies the inequality $\displaystyle f(x) \le A + B\int_a^x f(t)\, dt$, where $A$ and $B$ are constants with $B < 0$. If $\displaystyle \int_a^\infty f(x)\, dx$ exists, show that $\displaystyle \int_a^\infty f(x)\, dx \le -A/B$.

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[TeethWhitener]

My first instinct would be to simply take the limit as $x\to\infty$ of both sides of the inequality, on the assumption that if
$$\lim_{x\to\infty} \int_{a}^{x} f(t)dt$$
exists, then $\lim_{x\to\infty}f(x) = 0$. But once upon a time, @micromass taught me about functions like $f(x) = x^2 e^{-x^8\sin^2x}$, so I'm not sure that will work.

Edit: for folks who want to see what trickery is at work with this function:
http://www.wolframalpha.com/input/?i=plot+x^2e^{-x^8\sin^2(20x)}+from+0+to+3
The spikes get taller, thinner, and further apart as $x\to\infty$. This means that the function is integrable all the way out to infinity, even though $\lim_{x\to\infty} f(x)$ doesn't exist.

 

[Ackbach]

My first instinct would be to simply take the limit as $x\to\infty$ of both sides of the inequality, on the assumption that if
$$\lim_{x\to\infty} \int_{a}^{x} f(t)dt$$
exists, then $\lim_{x\to\infty}f(x) = 0$. But once upon a time, @micromass taught me about functions like $f(x) = x^2 e^{-x^8\sin^2x}$, so I'm not sure that will work.

This is Ackbach on Math Help Boards, and I'm the Director for the Undergraduate POTW there ( = Intermediate PotW here on PF).

That approach is not going to work, I agree. It's the first thought that occurred to me!

 

[TeethWhitener]

Alright, here's what I have so far.

Spoiler

Let
$$\int_{a}^{x} f(t)dt = F(x)$$
It follows that $F'(x) = f(x)$. The inequality becomes (with some rearrangement)
$$F'(x) - BF(x) - A \leq 0$$
In the case where the equality holds, we can solve the resulting differential equation to get:
$$F_{eq}(x) = -\frac{A}{B} + ce^{Bx}$$
We note that $\lim_{x\to\infty} F_{eq}(x) = -A/B$, or in other words:
$$\int_a^\infty f(x)\, dx = -\frac{A}{B}$$
In the case where the equality doesn't hold, we have
$$F'(x) - BF(x) - A = -\alpha$$
where $\alpha > 0$.
All this does is change the solution of the differential equation to
$$F(x) = -\frac{A-\alpha}{B} + ce^{Bx}$$
Taking the limit as $x\to\infty$ gives
$$\int_a^\infty f(x)\, dx = -\frac{A-\alpha}{B}$$
Since $B<0$, we have that
$$-\frac{A-\alpha}{B}<-\frac{A}{B}$$
or, combining the previous two results:
$$\int_a^\infty f(x)\, dx < -\frac{A}{B}$$
when the equality doesn't hold. Combining the two cases ($\alpha = 0$ and $\alpha \neq 0$) gives us the desired result:
$$\int_a^\infty f(x)\, dx \leq -\frac{A}{B}$$

 

[S.G. Janssens]

Nice problem.

Spoiler

Write $F(x) := \int_a^x{f(t)\,dt}$. In these terms, it was given that
$$
f(x) \le A + B F(x), \qquad \forall\,x \ge a.
$$
Integrate both sides from $a$ to $b > a$ and divide by $b - a$ to find
$$
\frac{F(b)}{b - a} \le A + \frac{B}{b - a}\int_a^b{F(x)\,dx}. \qquad (\ast)
$$
Since it was given that $F(\infty) := \lim_{b \to \infty}{F(b)}$ exists, it follows that
$$
\frac{1}{b - a}\int_a^b{F(x)\,dx} \to F(\infty), \qquad \text{as } b \to \infty
$$
and we have from $(\ast)$ that
$$
0 \le A + B F(\infty),
$$
and, using that $B < 0$, this yields $F(\infty) \le \tfrac{A}{-B}$.

(I don't want to enter any competition.)

 

[Ackbach]

Alright, here's what I have so far.

Spoiler

Let
$$\int_{a}^{x} f(t)dt = F(x)$$
It follows that $F'(x) = f(x)$. The inequality becomes (with some rearrangement)
$$F'(x) - BF(x) - A \leq 0$$
In the case where the equality holds, we can solve the resulting differential equation to get:
$$F_{eq}(x) = -\frac{A}{B} + ce^{Bx}$$
We note that $\lim_{x\to\infty} F_{eq}(x) = -A/B$, or in other words:
$$\int_a^\infty f(x)\, dx = -\frac{A}{B}$$
In the case where the equality doesn't hold, we have
$$F'(x) - BF(x) - A = -\alpha$$
where $\alpha > 0$.

Spoiler

How do you know that $\alpha$ doesn't depend on $x$? If it does, solving the DE would be impossible, because you wouldn't know the exact nature of the dependence. I'm not sure this approach works, either, I'm afraid.

 

[Ackbach]

Nice problem.

Spoiler

Write $F(x) := \int_a^x{f(t)\,dt}$. In these terms, it was given that
$$
f(x) \le A + B F(x), \qquad \forall\,x \ge a.
$$
Integrate both sides from $a$ to $b > a$ and divide by $b - a$ to find
$$
\frac{F(b)}{b - a} \le A + \frac{B}{b - a}\int_a^b{F(x)\,dx}. \qquad (\ast)
$$
Since it was given that $F(\infty) := \lim_{b \to \infty}{F(b)}$ exists, it follows that
$$
\frac{1}{b - a}\int_a^b{F(x)\,dx} \to F(\infty), \qquad \text{as } b \to \infty
$$

Spoiler

Why must that be? This step could use a bit more justification, I think.

 

[Ackbach]

For what it's worth, I have posted one correct solution on MHB.

Spoiler

http://mathhelpboards.com/potw-university-students-34/problem-week-280-sep-12-2017-a-22306.html

Here's another:

Spoiler

By Gronwall's inequality,
$$f(x)\le A\cdot \exp\left(\int_a^x B \, dt\right) = A \, e^{B(x-a)}.$$
Integrating both sides yields
$$\int_a^{\infty}f(x) \, dx \le \int_a^{\infty}A \, e^{B(x-a)} \, dx = -\frac{A}{B}$$
by the integral inequality. The result follows.

 

[TeethWhitener]

Spoiler

How do you know that $\alpha$ doesn't depend on $x$? If it does, solving the DE would be impossible, because you wouldn't know the exact nature of the dependence. I'm not sure this approach works, either, I'm afraid.

I don't think this is true.

Spoiler

Say $\alpha = \alpha(x)$. Then you have the inhomogeneous equation:
$$F'(x)-BF(x) = A-\alpha(x)$$
where $\alpha(x) > 0$ everywhere. This is solved with an integrating factor to give:
$$F(x)=-\frac{A}{B} + ce^{Bx}-e^{Bx}\int^x e^{-Bt}\alpha(t)dt$$
But since $\alpha(t)$ and $e^{whatever}$ are both positive everywhere within the domain of integration, the integral always subtracts a positive number from $-\frac{A}{B}$. So it's always less than $-\frac{A}{B}$, which is the desired result.

 

[Ackbach]

I don't think this is true.

Spoiler

Say $\alpha = \alpha(x)$. Then you have the inhomogeneous equation:
$$F'(x)-BF(x) = A-\alpha(x)$$
where $\alpha(x) > 0$ everywhere. This is solved with an integrating factor to give:
$$F(x)=-\frac{A}{B} + ce^{Bx}-e^{Bx}\int^x e^{-Bt}\alpha(t)dt$$
But since $\alpha(t)$ and $e^{whatever}$ are both positive everywhere within the domain of integration, the integral always subtracts a positive number from $-\frac{A}{B}$. So it's always less than $-\frac{A}{B}$, which is the desired result.

Fair enough! I'd accept this solution as correct.

 

[TeethWhitener]

Fair enough! I'd accept this solution as correct.

One other thing: In post #9, $\alpha(x) >0$ should be $\alpha(x) \geq 0$. But this doesn't change the conclusions.

 

[Ackbach]

One other thing: In post #9, $\alpha(x) >0$ should be $\alpha(x) \geq 0$. But this doesn't change the conclusions.

Spoiler

I'd agree. By the way, your approach in Post # 2 can be modified (as Opalg did in the posted solution on MHB) if you construct a clever subsequence $\{x_n\}$ such that $\{f(x_n)\}\to 0$.

 

[S.G. Janssens]

Why must that be? This step could use a bit more justification, I think.

It is because ordinary convergence of the improper integral (i.e. existence of $\lim_{b \to \infty}{F(b)}$ in $\mathbb{R}$) implies convergence (to the same limit) of the Cesàro means. As I do not have a reference for this ready at hand, here is an argument for the present context:

Spoiler

Let arbitrary $\epsilon > 0$ be given and let $x_{\epsilon} > 0$ be such that $|F(x) - F(\infty)| \le \tfrac{\epsilon}{3}$ for all $x \ge x_{\epsilon}$. Then, for $b \gt x_{\epsilon}$,
$$
\left|\frac{1}{b - a}\int_a^b{F(x)\,dx} - F(\infty)\right| \le \left|\frac{1}{b - a}\int_a^{x_{\epsilon}}{F(x)\,dx}\right| + \left|\frac{1}{b - a}\int_{x_{\epsilon}}^b{F(x)\,dx} - F(\infty) \right|. \qquad (\dagger)
$$
The first term in the RHS can be made less than $\tfrac{\epsilon}{3}$ by choosing $b$ sufficiently large. The second term equals
$$
\left|\frac{1}{b - a}\int_{x_{\epsilon}}^b{(F(x) - F(\infty))\,dx} - \frac{x_{\epsilon} - a}{b - a}F(\infty)\right| \le \frac{\epsilon}{3} + \left|\frac{x_{\epsilon} - a}{b - a}F(\infty)\right| \le \frac{\epsilon}{3} + \frac{\epsilon}{3},
$$
by choosing $b$ possibly even larger. In summary, by choosing $b$ large enough, one guarantees that the LHS of $(\dagger)$ does not exceed $\tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} = \epsilon$.

 

[S.G. Janssens]

Here's another:

Spoiler

By Gronwall's inequality,
$$f(x)\le A\cdot \exp\left(\int_a^x B \, dt\right) = A \, e^{B(x-a)}.$$
Integrating both sides yields
$$\int_a^{\infty}f(x) \, dx \le \int_a^{\infty}A \, e^{B(x-a)} \, dx = -\frac{A}{B}$$
by the integral inequality. The result follows.

In the problem it was assumed that $B < 0$, so I cannot see how Gronwall applies here. For example, let $a = 1$ and $f : [a,\infty) \to \mathbb{R}$ be given by $f(x) = \tfrac{1}{x^2}$. Then
$$
A + B \int_a^x{f(t)\,dt} = A + \frac{B}{a} - \frac{B}{x} \ge A + \frac{B}{a} = 1 \ge f(x), \qquad \forall\,x \ge a,
$$
if we choose $A = 2$ and $B = -1$. Moreover, the improper integral of $f$ exists. However, $A e^{B(x - a)} = 2 e^{-(x - 1)}$ for $x \ge a$ so
$$
f(x) > Ae^{B(x - a)}
$$
for $x$ large enough.

 

[Ackbach]

In the problem it was assumed that $B < 0$, so I cannot see how Gronwall applies here.

Hmm, you may be right. Looking into it...

 

[Ackbach]

You are correct that we can't use Gronwall directly. So my "solution" above is completely bogus. However, the proof method used for Gronwall can be adapted for this problem. Here's a note from my colleague Euge at MHB:

Spoiler

Set $\displaystyle g(x) = \int_a^x f(t)\, dt$, and note $g$ satisfies the differential inequality $g' - Bg \le A$, $g(a) = 0$. Multiply the inequality by $e^{-Bx}$ to obtain $\dfrac{d}{dx}(e^{-Bx}g) \le Ae^{-Bx}$, then integrate both sides from $x = a$ to $x = s$ (where $s$ is a fixed number $> a$) to get $e^{-Bs}g(s) \le -\dfrac{A}{B}(e^{-Bs} - e^{-Ba})$. Thus $g(s) \le -\frac{A}{B}(1 - e^{B(s-a)})$. Now let $s \to \infty$ to obtain the result.